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A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.

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A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 23 Aug 2015, 09:34
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A motor car travelling to point pedro overtakes a cyclist at 9:00A.M. The car reaches point pedro at 10:30 A.M. and after waiting for 1hour, returns meeting the cyclist at noon. When will the cyclist reach point pedro?

A. 1:15PM
B. 1:30PM
C. 1:45PM
D. 2:00PM
E: 1:28PM
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A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 22 Dec 2016, 07:34
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santorasantu wrote:
A motor car travelling to point pedro overtakes a cyclist at 9:00A.M. The car reaches point pedro at 10:30 A.M. and after waiting for 1hour, returns meeting the cyclist at noon. When will the cyclist reach point pedro?

A. 1:15PM
B. 1:30PM
C. 1:45PM
D. 2:00PM
E: 1:28PM


Hi
when it comes to time and speed and touching a point and coming back, it is better at times to break the time

A very easy and fast way to other wise a slightly complicated Q..

The motor car takes 9:00-10:30 or 1 and 1/2 hour to cover entire distance
And then starts back at 11:30 and meets the cyclist at 12:00, so meets after 1/2 hour...
So in that 1/2 hour it will cover 1/3 distance back as total time car takes is 1 and 1/2 hr..

So the cyclist would have covered 1-1/3= 2/3 distance in that time...
And it takes 12:00-9:00 or 3 hrs for it..
If 2/3 in 3 hrs, then complete route in 3*3/2=9/2 or 4 and 1/2 hr..
Time at that point =9:00+4 1/2 or 1330=1:30pm
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Re: A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 23 Aug 2015, 10:22
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actually this can be solved in less time than we would think it will take..

A - starting point where cyclist and motorist meet (9 AM)
B - meeting point after A, this happens when motorist returns (Noon)
C - Pedro - Motorist reaches here at 11.30 (1 hour of waiting).

We need to find time taken by cyclist to reach C from B.

Time taken for cyclist to reach B from A - 180 minutes (9 AM to Noon)
Time taken for Motorist to reach B from C - 30 mins (11.30 to Noon).
We know time taken by Motorist to reach C from A is 90 mins (9 to 10.30).
Hence time taken by motorist from A to B is 60 mins (AC-CB = 90-30 = 60)

so if it takes motorist 60 mins and 180 mins for cyclist for A to B, ratio of speed is 1:3,

Since motorist took 30 mins between C and B, it will take cyclist 90 mins (3 times). Hence answer is 90 mins after noon to reach C or 1.30 PM is the answer.

---------------------------------------

santorasantu wrote:
A motor car travelling to point pedro overtakes a cyclist at 9:00A.M. The car reaches point pedro at 10:30 A.M. and after waiting for 1hour, returns meeting the cyclist at noon. When will the cyclist reach point pedro?

A. 1:15PM
B. 1:30PM
C. 1:45PM
D. 2:00PM
E: 1:28PM


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A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 27 Dec 2016, 23:14
IMO B.

Car Travels entire distance in 1.5 hours. so it speed, if the distance is x, is 2/3x
While returning, it meets the cyclist at noon i.e. 12 i.e. after driving for 1/2 hour.
In 30 mins car will cover 1/3x distance. So in 3 hours cyclist would have covered 2/3x distance.
2/3 ---> 3 hours
1 ---> 4.5 hours -- 1.30PM
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Re: A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 28 Dec 2016, 03:35
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santorasantu wrote:
A motor car travelling to point pedro overtakes a cyclist at 9:00A.M. The car reaches point pedro at 10:30 A.M. and after waiting for 1hour, returns meeting the cyclist at noon. When will the cyclist reach point pedro?

A. 1:15PM
B. 1:30PM
C. 1:45PM
D. 2:00PM
E: 1:28PM


Make a diagram to understand:

..................9:00........................................................................10:30
Car ------->------------------------------------------>
Cyclist-----------------------------><----------------<
...............................................................12 noon......................11:30


From the point at 9:00, the car reaches pedro at 10:30 i.e. in 1.5 hrs.
From pedro to point at 12 noon, the car reaches in 0.5 hrs. So pedro to this point is 1/3rd distance.
So the cyclist covered 2/3 rd distance in 3 hrs. He will cover 1/3rd distance in half the time i.e. 1.5 hrs. So he reaches pedro at 1:30.
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A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 21 Feb 2018, 10:10
My approach:

For motor car, distance from cyclist to Point Pedro be 3X, time taken=1.5 hours. Therefore Speed of motor car= 2X(units)/hour.

Now, the motor car meets the cyclist after half an hour. Hence distance covered by motor car= speed*time= 2X*0.5= X.
At this point, the cyclist has already travelled, 3X-X= 2X. Therefore speed of cyclist= 2X/3(units)/hour (note he travelled from 9 A.M. to 12 P.M.).

Remaining distance for cyclist to cover= X. Time=?. Speed= 2X/3. Therefore, distance= speed*time; Time=Distance/Speed =X/(2X/3) = 3/2= 1.5 hours. Hence 12 + 1.5 hours= 1.30 P.M.

Hence, B.
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A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 21 Feb 2018, 10:27
b - speed by cyclist, c - speed of card , X - refers to cyclist
|<---1.5b---X------------------->| - car (distance = 1.5c) at 10:30
|<--------------2.5b--X-----------| - car (distance = 1.5c) at 11:30

now both meet at noon, so \(\frac{{1.5c - 2.5b}}{{b + c}} = 1/2\) => C = 3b
time taken to travel 1.5c - 2.5b is 4.5b - 2.5b = 2b , 2b/b = 2 hours => 11:30 + 2 = 1:30
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Re: A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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New post 22 Feb 2018, 17:46
santorasantu wrote:
A motor car travelling to point pedro overtakes a cyclist at 9:00A.M. The car reaches point pedro at 10:30 A.M. and after waiting for 1hour, returns meeting the cyclist at noon. When will the cyclist reach point pedro?

A. 1:15PM
B. 1:30PM
C. 1:45PM
D. 2:00PM
E: 1:28PM


We can let the distance between the place (where the car overtakes the cyclist) and Point Pedro = d. Also, we can let the car’s speed = r and cyclist’s speed = s. Thus, we have:

1.5r = d

And

3s + 0.5r = d

Multiply the second equation by 3 we have:

9s + 1.5r = 3d

Subtract the first equation from this, we have

9s = 2d

4.5s = d

Thus it takes 4.5 hours for the cyclist to reach Point Pedro from the place (where he is overtaken by the car). Thus he reaches Point Pedro at 9:00 AM + 4.5 hours = 13:30 = 1:30 PM.

Answer: B
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Re: A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.  [#permalink]

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Re: A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.   [#permalink] 26 Aug 2019, 02:29
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